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| 1 | +package DynamicProgramming; |
| 2 | + |
| 3 | + |
| 4 | +import java.util.Arrays; |
| 5 | + |
| 6 | +/** |
| 7 | + * Recursive Solution for 0-1 knapsack with memoization |
| 8 | + */ |
| 9 | +public class KnapsackMemoization { |
| 10 | + |
| 11 | + private static int[][] t; |
| 12 | + |
| 13 | + |
| 14 | + // Returns the maximum value that can |
| 15 | + // be put in a knapsack of capacity W |
| 16 | + public static int knapsack(int[] wt, int[] value, int W, int n) { |
| 17 | + if(t[n][W] != -1) { |
| 18 | + return t[n][W]; |
| 19 | + } |
| 20 | + if (n == 0 || W == 0) { |
| 21 | + return 0; |
| 22 | + } |
| 23 | + if (wt[n - 1] <= W) { |
| 24 | + t[n-1][W-wt[n-1]] = knapsack(wt, value, W - wt[n - 1], n - 1); |
| 25 | + // Include item in the bag. In that case add the value of the item and call for the remaining items |
| 26 | + int tmp1 = value[n - 1] + t[n-1][W-wt[n-1]]; |
| 27 | + // Don't include the nth item in the bag anl call for remaining item without reducing the weight |
| 28 | + int tmp2 = knapsack(wt, value, W, n - 1); |
| 29 | + t[n-1][W] = tmp2; |
| 30 | + // include the larger one |
| 31 | + int tmp = tmp1 > tmp2 ? tmp1 : tmp2; |
| 32 | + t[n][W] = tmp; |
| 33 | + return tmp; |
| 34 | + // If Weight for the item is more than the desired weight then don't include it |
| 35 | + // Call for rest of the n-1 items |
| 36 | + } else if (wt[n - 1] > W) { |
| 37 | + t[n][W] = knapsack(wt, value, W, n - 1); |
| 38 | + return t[n][W]; |
| 39 | + } |
| 40 | + return -1; |
| 41 | + } |
| 42 | + |
| 43 | + // Driver code |
| 44 | + public static void main(String args[]) { |
| 45 | + int[] wt = {1, 3, 4, 5}; |
| 46 | + int[] value = {1, 4, 5, 7}; |
| 47 | + int W = 10; |
| 48 | + t = new int[wt.length+1][W+1]; |
| 49 | + Arrays.stream(t).forEach(a -> Arrays.fill(a, -1)); |
| 50 | + int res = knapsack(wt, value, W, wt.length); |
| 51 | + System.out.println("Maximum knapsack value " + res); |
| 52 | + } |
| 53 | +} |
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